A car exciting the highway moving at 27 m/s slows down at the rate of 3.5 m/s^2 to a final velocity of 18 m/s. Calculate the distance the car travels while braking?
the velocity changes by 9 m/s (27 - 18)
this takes approx. 2.6 s (9 / 3.5)
the average velocity is 22.5 m/s
___ (27 + 18) / 2
the distance traveled is velocity multiplied by time
To calculate the distance the car travels while braking, we can use the equation of motion:
\(v_f^2 = v_i^2 + 2a d\)
where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(d\) is the distance traveled.
In this case, the car's initial velocity is \(v_i = 27 m/s\), the final velocity is \(v_f = 18 m/s\), and the acceleration is \(a = -3.5 m/s^2\) (negative because the car is slowing down).
Rearranging the equation to solve for \(d\), we have:
\(d = \frac{{v_f^2 - v_i^2}}{{2a}}\)
Substituting the given values, we get:
\(d = \frac{{18^2 - 27^2}}{{2 \times -3.5}}\)
Evaluating the expression:
\(d = \frac{{324 - 729}}{{-7}}\)
\(d = \frac{{-405}}{{-7}}\)
\(d = 57.86\) meters (rounded to two decimal places)
Therefore, the car travels approximately 57.86 meters while braking.